The parameters a and b determine how close the
solution starts to the steady state solution. Let us take a = 0.4 and b = 1.8. To determine the steady state solution,
we find from the equation y2¢(t) = 0 that when m = 0,
y2,0 = (y1,0 + a)/b. Using this in the equation y1¢(t) = 0 for m = 0, we find that y1,0 satisfies the algebraic
equation

-b y3 + (3 b + 1) y + 3 a = 0 .

After computing all the roots of this cubic equation with
roots, y1,0 is the unique real root bigger than 1 -r b. For the given a,b this results in y1,0 = 2.417960226013935. Using this value, compute y2,0 and solve
the problem for t = 20 and various m, say m = 1, -1, 10,-10. The figures for two of these values show what you might
find.